What is the integral from 0 to 1 of x/((3x^2+1)^4) ?

The integral from 0 to 1 of x/((3x^2+1)^4) is 7/128

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integral from 0 to 1 of x/((3x^2+1)^4)

\int_{0}^{1} \frac{x}{( 3 x ^{2} + 1 ) ^{4}} d x

\frac{7}{128}

Decimal

0.0546875

Solution steps

\int_{0}^{1} \frac{x}{( 3 x ^{2} + 1 ) ^{4}} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{6 u ^{4}} d u

= \int_{1}^{4} \frac{1}{6 u ^{4}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{1}^{4} \frac{1}{u ^{4}} d u

Apply the Power Rule

: [- \frac{1}{3 u ^{3}}]_{1}^{4}

= \frac{1}{6} [- \frac{1}{3 u ^{3}}]_{1}^{4}

Compute the boundaries:

\frac{21}{64}

= \frac{1}{6} \cdot \frac{21}{64}

Simplify

= \frac{7}{128}

\int_{0}^{2 x} \frac{x y}{x ^{2} + y ^{2}} d y

\int_{- 3}^{0} x^{2} d x

x \to 3 lim (+ (\frac{- 3}{x - 3}))

\frac{d ^{2}}{d x ^{2}} (- 9.8 x^{2} + 2 x + 1)

e^{t} (7 y - 5 t) d t + (6 + 7 e^{t}) d y = 0

What is the integral from 0 to 1 of x/((3x^2+1)^4) ?
The integral from 0 to 1 of x/((3x^2+1)^4) is 7/128